Are you saying that you now feel that you didn't understand the tide effect when you wrote the OP? I had thought it summed up the BA's position pretty well--other than the other questions you had about Mercury.

Dr.Plait! I wondered if the conversation had got to a point when we needed to appeal to a Higher Power, but that of course would be against everything you stand for. Nevertheless, please take part (no one else is!).

Eroica quotes an explanation that invokes centrifugal force - that Great Chimera - but, Yes, that is what I mean. Swing a weight around on a string, and you feel a pull. Swing it faster and the pull gets stronger. The weight is going faster than it needs to stay in "orbit" about your finger and so tends to seek a higher orbit where at that speed it will be stable. The converse is difficult to simulate in this way, and the whole idea can only be an analog, as a string doesn't work like gravity.

I am flattered by Eroica's epiphany, and I can share her feeling. I've had the same moment, for instance when trigonometry became 'true' for me - it involved a 'vision' of a right angled triangle that changed its size, angles and sides, but always in proportion! But away from solipsism.

I think we have evolved a test. Sowicki used his model to calculate a value for the force that raises the tides. Can that force be calculated from the "different orbits"(DO) model? If the DO value agrees with Sawicki's value, then the two models are probably correct but different views of the same reality.

I regret that I would have to go back to school to learn how to do so. Eroica, if you are able, please will you give it a try? Dr.Plait? Kilopi? All? What I tell you three times is true!
John

I've never said your model was not correct--but there are some gotchas along the way.

The main one, as near as I can tell (the one that seems to trip up even the pros), is using a rotating Earth to calculate the tide. You don't have to ignore reality, but if you do the calculation with the Earth rotating, then you have to account for the effect of the Earth rotating. That's why it appears as a complication-- but that doesn't make the theory wrong.

The second one, which seems to surprise a lot of people, is that the centrifugal force caused by the revolution is exactly the same at every point on the Earth. Any difference is caused by the rotation of the Earth.

There are a lot of ways to explain the concept--but that doesn't mean the BA's way is wrong, or worse than any other.

All,
I was reviewing this thread, and saw what I had missed before - Kilopi referring to a posting by Dr.Sawicki. Following that thread to its origin, I found a posting by JohnWitt, expressing precisely the "different orbits" point of view, though he spoils his pitch by mentioning centrifugal force!

John, are you out there? Or even Dr.Sawicki? Come and join us! It is good to know that there are others as well as Eroica and myself who see the Universe this way, and I would welcome more who can test the two points of view.
John

You can add me to the list if you like.

Funny. I thought my OP expressed a completely different and completely wrong view (and one which I now retract)! Originally I thought that the tidal bulge on the far side was being "left behind" because the rest of the Earth was being pulled towards the Moon more strongly. But I now realize that if this were literally happening, the Earth would soon collide with the Moon! The Earth is not "lurching towards" the Moon. The CoM of a planet in a circular orbit about a star remains at the same distance from the barycentre. (In fact, over time it moves away; but let's not complicate things!)

But if there is something in the "different orbits" model, I still don't understand why the rotation of the Earth is not a factor. After all, we launch rockets from as close to the equator as possible because their initial velocity due to the Earth's rotation plays a very real part in helping them achieve escape velocity. Why should the same not be true for a mass of water on the far side of the Earth as it tries to 'climb' to a higher orbit? How does its velocity due to the Earth's revolution differ from its velocity due to the Earth's rotation (other than in magnitude)?

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I'm not sure if the word "lurching" was used before, but it is true that the Earth can be thought of as accelerating towards the Moon. If you want to call that lurching, then I don't see a problem with that point of view either.
As the Earth rotates, its equator bulges. If you only look at the near side and the far side (from the moon), you'd notice that they both had a bulge because of the rotation. It is tempting to attribute those bulges to the tides.

But, if you look at other places along the equator, they bulge also! There is no tidal depression. And the rotational bulge is more or less permanent. It doesn't change with time.

Plus, the tidal bulges are near the ecliptic, whereas the bulges due to rotation are aligned along the equator. (The ecliptic--the path of the Sun from our point of view-- is tilted over twenty degrees with respect to the equator, and the plane of the moon's orbit is much closer to the ecliptic than to the equator)

So the rotation of the Earth does cause a bulge, but not what we'd normally call a tidal bulge. The rotation of the Earth has produced a bulge about the equator that is twenty kilometers farther from the center of the Earth than the poles.

PS: I just checked, and "lurch" was used in the OP.

Not to me it ain't. I'm not talking about or thinking of the rotational bulge, since it has nothing to do with the Sun or the Moon. That's why I introduced the idealized model of a planet in locked rotation, where the rotation is so slow that it can be discounted.

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Sawicki's figure is the difference between the acceleration due to the Sun's gravity at the centre and at the near (or far) point on the Earth's surface. My own calculations led to a figure of 1.01e-6 metres per second-squared for the difference between the near and far points. That's actually the same as Sawicki's figure (if you double it and replace g with 9.81 metres per second-squared). But all that this means is that I can still do arithmetic.

To reconcile your theory with Sawicki's you need to calculate the sizes of the bulges that would be raised, and at the moment I don't know how to do that. I fear you are overestimating my abilities. I think we need reinforcements.

(PS: Eroica is a guy.)

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Boy, these tides are starting to do my head in!

JohnD, I have been thinking about your first post to this thread. At the time I criticised it because you said that the Earth's CoM was in orbit about the Moon, whereas it is actually in orbit about the barycentre.

Now, a question occurs to me: why are we talking about the Earth-Moon barycentre at all, when we are trying to explain lunar tides? The barycentre is the point where the Moon and the Earth's gravity are centred. But the Earth's gravity plays no role in creating the lunar tides (other than preventing the lunar tides from tearing the Earth apart).

Maybe JohnD was right. We should be considering the Earth's motion relative to the Moon's CoM! I'll try crunching some numbers when I get a chance.

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But why not go all the way, and get rid of the rotation entirely? That way, you can be sure it can be discounted.
What's hard for me to understand is why you would say the BA's approach is hard to understand, yet we don't have a calculation for the alternative theory. My goal is to convince you that that approach is not as simple as it may seem.

I've been thinking about the barycentre again, and I now think that Sawicki and the BA are wrong when they claim that a person at the Earth's CoM doesn't feel the force of the Moon's gravity because they are in free fall about the barycentre. They are in free fall about the barycentre, but that's not why they don't feel the Moon's gravity. The correct reason is that they are also in free fall about the Moon's CoM!

I haven't done (and perhaps can't do) the maths, but Starry Night Pro clearly demonstrates that if you take the Moon's CoM as your fixed frame of reference, then the Earth's CoM does indeed revolve about it in a neat elliptic path, and is therefore in free fall (like all objects travelling through a gravitational field along a path that is a conic section).

If this analysis is correct, then my objection to JohnD's first post is no longer valid. The fact that the barycentre is between the tidal bulges is irrelevant to lunar tides. What is relevant is that they are both on the same side of the Moon's CoM.

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Eroica,
Apologies for mistaking your gender - Eroica is a fine 'handle' but would be a splendid and original first name for a girl! Though I suppose adding a 't' could cause all sorts of embarassment.

I was delighted to read that you had calculated the same value for delta-a(?) as Sawicki, but then realised you were doing it from his model; Yes? So proves nothing, except as you say your arithmetic is reliable. But why is it necessary to calculate the size of the bulge? That follows from the value of the force.

I'm happy to be considering the barycentre rather than the CoM of the Moon. My original mind-picture was of a small satellite around a large primary, Earth-space probe or Sun-Earth, so that the barycentre virtually coincided with the primary's CoM, and I had not visualised it. You corrected that. the model still works and it is the different orbits of parts of the satellite around the barycentre that lift the tides, not orbits around the CoM of the other partner in the binary

Kilopi,
Are you getting a bit hung up on the equatorial bulge? That is real, but static, caused as you say by the Earth's rotation. And I'm happy to accept that it is of the order of 20 kilometers - I can't dispute it! But we are discussing tides that nowhere on Earth vary by more than 50ft/16m, an ocillation of less than 0.1% compared to the bulge you quote. Surely such an ocillation can be superimposed on the equatorial bulge?

Which raises again my own ignorance - Eroica (did I see Dublin associated with your name) and I live in mid latitudes, where tides are significant. Do tides occur around the Equator? I had always assumed that they did, but are you implying, Kilopi, that the bulge somehow suppresses them, so they are much less high, or absent?
And then what about the Poles. Apart from axial tilt, the Poles have the same orbit around the primary as the CoM, so no tidal effects there - Sawicki points this out from his model, and the DO model does the same. (Another parallel!) Following this point back to the start, BOTH models predict that tidal forces will be maximal at the Equator (axial tilt excepted) Unless you can tell me that there are no, or reduced tides on the Equator, I'll insist on wobbling the bulge!

John

Any point of view can be used, not just a luna-centric one. The math does get complicated sometimes though.
Hung up? Just the opposite. My warning is that one should make sure that the calculations don't accidentally mistake a rotational component as a tidal component. If your model includes a rotation of the Earth, you have to somehow subtract that out to calculate the tidal effect.
That's my whole point.
?? No, to the last question, yes to the first.

I used the idea of subtracting vectors because it's easier to swallow, for me, than one using centrifugal force. I have no problems with centrifugal force, but it implies a rotating system. You get tides even if the system is not rotating. Let the Earth fall into the Sun, and you still get differential tides across the Earth, even when the net angular momentum is zero. So using centrifugal force won't apply there.

I also wanted to use a different explanation than is usually seen, because why repeat something in every textbook? Actually, I have gone through many textbooks, and very few explain tides in any detail at all, a fact I find astonishing. Tides sculpt many objects in the Universe, from our shorelines to clusters of galaxies. It's worth spending a few hundred words explaining carefully!

I sure hope you do, BA, as it needs someone to handle it properly. Too bad it's not simple, however.

For instance, the gravitational force from F=G*M1*M2/r^2 between me (90kg) and the moon says I am only .0007 lbs. lighter when the moon is overhead. Surprisingly, this same force equation says I am .12 lbs lighter when the sun is overhead. So the Sun's gravitational force is 170x's greater than the moon.

The websites show that tidal forces, however, vary as the inverse cube of the distance not the square. I wish I could find a clear explanation for this.

Also, it appears the horizontal component not the radial component is the main player in an ocean tide.

Here is an animated vector tidal force site that is pretty neat....

>>> Round n Round We Go <<<

For fun, and to use exaggeration to emphasize gravity forces, I used the gravitational equation for a blackhole. If I get within 1/2 million miles of a million sun BH, the force on me becomes 4 billion lbf. The only difference in force for 1 meter closer is only about 10 lbf. However, at 100,000 miles this difference in 1 meter is about 6,000 lbf. Ouch!

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I wish I had never clouded the issue by talking about the Earth's rotation or by linking to that MadSci article. So let me clarify one thing: the DO (Different Orbits) theory does not - NOT involve centrifugal forces. If anything, the BA's "negative force" is closer to the notion of centrifugal forces than anything in the DO theory. The only force in the DO theory is gravity, which is always attractive. The DO theory tries to account for the tidal bulge's movement away from the Moon in terms of its velocity through the Moon's gravitational field.

So let's simplify things and ignore completely the Earth's rotation. If necessary I'll start a new thread some day to address that issue. kilopi challenged me to describe the DO with a non-rotating Earth. I'll try.

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As far as I see there is no real difference at all between the DO theory and the DF (Differential Forces) theory in the case where a non-rotating Earth is simply falling into the Sun. In that case, the tidal bulge on the far side really is being left behind by the rest of the Earth, while the bulge on the near side is outstripping the rest of the planet. It's the way I envisaged the tides in my OP.

But imagine, for a moment, what would happen in such a case if the Earth were prevented from falling. For example, imagine there is a spike driven through the centre of the Earth and protruding from the poles; now imagine that a giant being took hold of the ends of that spike and held the Earth above the Sun. What sort of tidal bulges would you get? There would still be a bulge on the near side as the Earth tried to slip off the spike and into the Sun, but there would be no bulge on the far side. The Earth would be pear-shaped. It would be like an ice-cream lolly that is being held horizontally to the ground, and is starting to melt and slide off the stick!

In this (admittedly fantastic) scenario, the far side of the Earth does not bulge away from the Sun because it is not moving through the Sun's gravitational field with the necessary velocity to climb up the gravitational gradient. In fact, it's not moving at all. It's like a rocket sitting on the launch-pad: it ain't going nowhere!

But those same differential forces that produced the normal tidal bulges still apply. In this case they don't produce the normal tides because the Earth's CoM is no longer in free fall. A person at the centre of the Earth would feel the Sun's gravity. But someone on the far side of the Earth would still feel a weaker force than someone at the CoM!

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What goes around comes around, this seems to me a rather more
eloquent version of the debate I had here http://www.badastronomy.com/phpBB/viewtopic.php?t=887.

I am still not altogether happy with the books explanation, but I did gain a much better understanding of tides largely thanks to GrapesofWrath.

If I might chip in a few points.

George wrote:
The key to tides is not the absolute force, but the relative force. Try redoing the calculation for the force exerted on you at midday and midnight.

Kilopi said
The cetrifictional force (a name a hadn't heard before ) is constant, but it seems to me something of a stretch to say it can now be ignored.

I think another wild analogy is called for. Imagine the earth were made of some material whose stiffness were controlled by an electric current. So if we start in the high stiffness mode, we have small tidal bulges. If we flip a switch and suddenly the Earth is less stiff, the tidal bulges increase. Part of the Earth moves away from the moon. I think the best way to explain this is the cetrifictional force.


Simon

P.S. I feel the need to point out that I don't just come here to argue about tides. I haven't visited in a while, and this was at the top of the board.

The one that I found helpful was based on the idea in BA's book. The tidal force is a result in the difference between the gravitational forces at two different places. In other words, it's more or less the derivative of the gravitational force. Take a derivative of the inverse square, and you have an inverse cube.
I'm not sure exactly what you are referring to here. I've seen some similar discussion, but I'm not clear on what you mean by "horizontal." Could you go into more detail?

Good idea. My body will experience a difference between these two forces of 0.000067 lbf. [Lunar attraction only]

So, the simple gravitational attraction of the moon seems far too weak to explain tides.

I stumbled into this site that brings up a very interesting idea.

>>> here <<< .

It says that tides are more of a standing wave. The bodies of waters are like oversize tanks of water that if you shake at a resonant frequency you get bigger and bigger waves (tides) in the tank.

Possibly my only outstanding contribution to mankind, as a youth at least, may apply here. I can get into almost any swimming pool at about 4 feet depth and oscillate my body in a certain way and at a certain frequency, and in little time, create white caps on the enlarging waves generated. 8) =D> *chest swelling* (no autographs please).

I suppose the "shaking" force would be the Lunar gravity force vector component that acts along the east-west plane of the sea. The resonant frequency, of course, is critical to the ocean depth so this helps explain the significant differences in local tides.

So ya'll's thoughts on this would be appreciated (ocean resonance, that is not my pool ability).

BTW - my wife can tie a cherry stem into a knot with just her tongue. 8) =D> [No, it's not why I married her]

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Yep. It just didn't seem intuitive at first but it does with a little thought as there can be little difference my body feels as a result of the Sun's extensive distance.

Good question. "Horizontal" is a bit akward term for a vector component

The "pull" of the Moon is vertical at only one point on the Earth's surface at any given time. All other points will be "pulled" at an angle. This force at an angle (vector) can be broken into two components - vertical and horizontal to the Earth's surface. The further from that vertical point, the greater the horizontal force. However, it won't be as large as the wimpy vertical component as the angle is never that great. (Of course, you knew all this but I figure someone might not).

So how can it help make a tide? It may be very important if the tide is a standing wave and this component keeps building the wave up along with other forces like the sun.

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All,
As I said at the start, I honestly believed that the different orbits and the gravity vector models were compatible and described the same situation in different ways. However, I've been thinking hard about the Earth free-falling into the heart of the Sun, and in this situation I can't explain tides using the DO model. So by Occam's Razor, the DO model fails, the GV wins, and I retire from the discussion. Sadly, as I still find the GV model difficult to understand, Thankfully, as the discussion has been most enjoyable!
John

The actual tides vary all over the place. As has been mentioned before, the force which produces the tides would result in less than a meter difference from high tide to low tide, but places like the Bay of Fundy, where the tidal range is on the order of a dozen meters, are infamous.
We all do.
Cheers!

I can. And I think I did in my last post.
Of course, in this situation, every point on (and in) the Earth is orbiting too slowly to stay in orbit and so they all fall into lower orbits. The difference is that their orbital speeds are not just too slow, they're zero.

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Thanks for that site. It verifys my point that resonance plays a key role.
As per your site...

I do not wish to diminish the DO - GV arguments, as they are stimulating, but the resonant factor explains much.

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There is nothing wrong with the DO model per se--but it should be noted that the velocity of every point on the Earth is essentially identical, in magnitude and direction, disregarding rotation. Thus, the velocities do not induce a differential from farside to center to nearside.

Still, the model may be instructive for some. That doesn't make the BA's not as good.

[You probably are wore out already, but you did answer my threads and I appreciate it].

I'm sure the simple answer is that it's ratio of it's rotation (58.65 days) to it's revolution (88 days) is a nice 3/2 harmonic motion. It is "comfortable" there like a vibrating string.

Now, if you want to know why it is not 1/1 (like a vibrating string prefers) well, this is for someone wiser. Most moons are 1/1, I believe. Maybe planetary surface characteristics such as bulges, mountains and water play a stronger tidal roll. It wouldn't surprise me to see our friend "General Relativity" step in here in the case of Mercury.

BTW - If you don't mind, how do you say your name? Mine is obvious but I have not heard yours before. Is it Eh-Roy-Ka, or what?

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Since posting that first message, I have done a bit of googling and discovered that the primary reason that Mercury is locked into a 3:2 resonance, rather than a 1:1 like our Moon, is because its orbit is much more eccentric (0.2056) than the Moon's (0.0549). If it were locked into a 1:1 resonance, it would not be able to keep the same side facing the Sun because its orbital velocity would fluctuate a good deal between aphelion and perihelion, whereas its rotational velocity would be more or less unchanging.

Since it can't easily rotate at different speeds, there's no way for Mercury to keep the same face, and its tidal bulge, pointed at the sun at all times.

Instead, it does the best it can: it makes sure that it always has its tidal bulges lined up with the sun when it is closest to the Sun and the force of gravity is strongest. The best way to do that while rotating at a constant speed is to complete 1.5 rotations every year. That way the tidal bulges always line up with the sun at perihelion.

Our Moon's orbit is also elliptic, so its orbital speed fluctuates, but not as much. Nevertheless, these fluctuations do cause longitudinal librations, allowing us to see a little more than 50% of the Moon's face. (There are other causes for the various other forms of libration.)

PS: Eroica is the Italian word for "Heroic." I borrowed it from my favourite piece of music, Beethoven's third symphony, the Sinfonia Eroica. I pronounce it er-ro-ik-ka. I've been using it as a nickname since long before the Internet.

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The tidal bulge is always pointed at the Sun, except for the slight offset that results in tidal slowing. If Mercury is in a resonance, the slowing on one part of the orbit must be offset by speeding up on another part.
There may be density differences that were emplaced a long time ago that have nothing to do with the tides that tend to orient the planet. Perhaps they were a result of the impact that produced the Caloris basin.